Hi please help me I\'m having a really hard time I asked a question before but i

Question

Hi please help me I'm having a really hard time I asked a question before but it wasn't help full. If someone can go into detail with the numbers and answers please. Thank you



http://www.physics.qc.edu/files?455

Suppose the volume of the syringe is 40.6 cc. You measure the pressure as 1.84 atm, and the temperature as 23.4o C. Assume: the diameter of the plunger on the syringe is 3.05 cm.

a)Find
- n, the number of moles of gas in the syringe: ---- moles
- N, the number of molecules of gas in the syringe: ----- molecules
- Use the molear mass of air given to calculate the mass of air in the syringe: ---- kg

b)Supose you push down the syringe so the volume is now 34.1 cc, with the temperature remaining unchanged. Find:
the new pressure inside the syringe. ----atm
the force exerted by the gas on the plunger of the syringe: ------ N


a) Assume that air pressure is 1.00 atm and the air temperature is 19.9o C. If the abolute zero apparatus is open to the air, how many moles of air will it hold?
----moles

b) Suppose now that the apparatus is sealed, and the temperature rises to 63.9o C. Find the pressure inside the apparatus now.
----atm

Explanation / Answer

(a.) n=pV/RT=1.85*1.01*10^5*40.6*10^-6/8.314/(273+23.4)=3.078*10^-3 N=6.023*10^23*n=1.853*10^21 m=nM=3.078*10^-3*28.966=0.089 grams (b.) Since no. of moles is also same,so p1V1=p2V2 so p2=p1V1/V2=1.84*40.6/34.1=2.19 atm Force=pressure * area=p2*pi*d^2/4=161.6 N (a.) Volume of sphere=4*pi*r^3/3=548.857*10^-6 m^3 n=pV/RT=1.01*10^5*548.857*10^-6/8.314/(273+19.9)=0.0227 moles p1/T1=p2/T2 so p2=1.15 atm

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.